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2026-01-01
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>We use the derivative of 6/x, which is -6/x², as a tool to understand how the function changes with respect to x. Derivatives help us analyze various real-world scenarios involving rates of change. We will now discuss the derivative of 6/x in detail.</p>
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<p>We use the derivative of 6/x, which is -6/x², as a tool to understand how the function changes with respect to x. Derivatives help us analyze various real-world scenarios involving rates of change. We will now discuss the derivative of 6/x in detail.</p>
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<h2>What is the Derivative of 6/x?</h2>
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<h2>What is the Derivative of 6/x?</h2>
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<p>We now understand the derivative of 6/x. It is commonly represented as d/dx (6/x) or (6/x)', and its value is -6/x². The<a>function</a>6/x has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>We now understand the derivative of 6/x. It is commonly represented as d/dx (6/x) or (6/x)', and its value is -6/x². The<a>function</a>6/x has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>The key concepts are mentioned below: </p>
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<p>The key concepts are mentioned below: </p>
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<ul><li>Function Representation: (6/x is equivalent to 6·x⁻¹). </li>
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<ul><li>Function Representation: (6/x is equivalent to 6·x⁻¹). </li>
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<li>Power Rule: Rule for differentiating functions of the form axⁿ. </li>
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<li>Power Rule: Rule for differentiating functions of the form axⁿ. </li>
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<li>Negative Exponents: Understanding that x⁻¹ is equivalent to 1/x.</li>
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<li>Negative Exponents: Understanding that x⁻¹ is equivalent to 1/x.</li>
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</ul><h2>Derivative of 6/x Formula</h2>
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</ul><h2>Derivative of 6/x Formula</h2>
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<p>The derivative of 6/x can be denoted as d/dx (6/x) or (6/x)'.The<a>formula</a>we use to differentiate 6/x is: d/dx (6/x) = -6/x² (or) (6/x)' = -6/x²</p>
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<p>The derivative of 6/x can be denoted as d/dx (6/x) or (6/x)'.The<a>formula</a>we use to differentiate 6/x is: d/dx (6/x) = -6/x² (or) (6/x)' = -6/x²</p>
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<p>The formula applies to all x where x ≠ 0.</p>
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<p>The formula applies to all x where x ≠ 0.</p>
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<h2>Proofs of the Derivative of 6/x</h2>
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<h2>Proofs of the Derivative of 6/x</h2>
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<p>We can derive the derivative of 6/x using proofs. To show this, we will use algebraic manipulation along with the rules of differentiation.</p>
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<p>We can derive the derivative of 6/x using proofs. To show this, we will use algebraic manipulation along with the rules of differentiation.</p>
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<p>There are several methods we use to prove this, such as:</p>
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<p>There are several methods we use to prove this, such as:</p>
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<ul><li>By First Principle </li>
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<ul><li>By First Principle </li>
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<li>Using Power Rule</li>
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<li>Using Power Rule</li>
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</ul><h2><strong>By First Principle</strong></h2>
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</ul><h2><strong>By First Principle</strong></h2>
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<p>The derivative of 6/x can be proven using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 6/x using the first principle, we will consider f(x) = 6/x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 6/x, we write f(x + h) = 6/(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [6/(x + h) - 6/x] / h = limₕ→₀ [(6x - 6(x + h)) / (x(x + h))] / h = limₕ→₀ [-6h / (x² + xh)] / h = limₕ→₀ [-6 / (x² + xh)] As h approaches 0, the<a>expression</a>becomes: f'(x) = -6/x² Hence, proved.</p>
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<p>The derivative of 6/x can be proven using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 6/x using the first principle, we will consider f(x) = 6/x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 6/x, we write f(x + h) = 6/(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [6/(x + h) - 6/x] / h = limₕ→₀ [(6x - 6(x + h)) / (x(x + h))] / h = limₕ→₀ [-6h / (x² + xh)] / h = limₕ→₀ [-6 / (x² + xh)] As h approaches 0, the<a>expression</a>becomes: f'(x) = -6/x² Hence, proved.</p>
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<h2><strong>Using Power Rule</strong></h2>
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<h2><strong>Using Power Rule</strong></h2>
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<p>To prove the differentiation of 6/x using the<a>power</a>rule, We rewrite the function as 6·x⁻¹ and apply the power rule: d/dx [6·x⁻¹] = 6·(-1)x⁻² = -6/x² Thus, the derivative of 6/x is -6/x².</p>
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<p>To prove the differentiation of 6/x using the<a>power</a>rule, We rewrite the function as 6·x⁻¹ and apply the power rule: d/dx [6·x⁻¹] = 6·(-1)x⁻² = -6/x² Thus, the derivative of 6/x is -6/x².</p>
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<h2>Higher-Order Derivatives of 6/x</h2>
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<h2>Higher-Order Derivatives of 6/x</h2>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a bit complex. To understand them better, consider a scenario where certain quantities change at different rates. Higher-order derivatives help analyze such functions effectively.</p>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a bit complex. To understand them better, consider a scenario where certain quantities change at different rates. Higher-order derivatives help analyze such functions effectively.</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>For the nth Derivative of 6/x, we generally use fⁿ(x) for the nth derivative of a function f(x) to understand the<a>rate</a>of change at different levels.</p>
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<p>For the nth Derivative of 6/x, we generally use fⁿ(x) for the nth derivative of a function f(x) to understand the<a>rate</a>of change at different levels.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When x = 0, the derivative is undefined because 6/x has a vertical asymptote there.</p>
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<p>When x = 0, the derivative is undefined because 6/x has a vertical asymptote there.</p>
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<p>When x = 1, the derivative of 6/x = -6/(1)², which is -6.</p>
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<p>When x = 1, the derivative of 6/x = -6/(1)², which is -6.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 6/x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 6/x</h2>
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<p>Students frequently make mistakes when differentiating 6/x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
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<p>Students frequently make mistakes when differentiating 6/x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
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<h3>Problem 1</h3>
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<h3>Problem 1</h3>
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<p>Calculate the derivative of (6/x)·(x²)</p>
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<p>Calculate the derivative of (6/x)·(x²)</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(x) = (6/x)·(x²). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 6/x and v = x². Let’s differentiate each term, u′ = d/dx (6/x) = -6/x² v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = (-6/x²)·(x²) + (6/x)·(2x) Simplifying terms, f'(x) = -6 + 12 = 6 Thus, the derivative of the specified function is 6.</p>
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<p>Here, we have f(x) = (6/x)·(x²). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 6/x and v = x². Let’s differentiate each term, u′ = d/dx (6/x) = -6/x² v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = (-6/x²)·(x²) + (6/x)·(2x) Simplifying terms, f'(x) = -6 + 12 = 6 Thus, the derivative of the specified function is 6.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the derivative of the given function by dividing the function into two parts.</p>
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<p>We find the derivative of the given function by dividing the function into two parts.</p>
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<p>The first step is finding its derivative and then combining them using the product rule to get the final result.</p>
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<p>The first step is finding its derivative and then combining them using the product rule to get the final result.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>A company finds that the cost of production is represented by the function y = 6/x where y represents the cost at production level x. If x = 3 units, calculate the rate of change of cost.</p>
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<p>A company finds that the cost of production is represented by the function y = 6/x where y represents the cost at production level x. If x = 3 units, calculate the rate of change of cost.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>We have y = 6/x (cost function)...(1) Now, we will differentiate the equation (1). Take the derivative of 6/x: dy/dx = -6/x² Given x = 3, substitute this into the derivative: dy/dx = -6/(3)² = -6/9 = -2/3 Hence, the rate of change of cost at a production level of x = 3 units is -2/3.</p>
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<p>We have y = 6/x (cost function)...(1) Now, we will differentiate the equation (1). Take the derivative of 6/x: dy/dx = -6/x² Given x = 3, substitute this into the derivative: dy/dx = -6/(3)² = -6/9 = -2/3 Hence, the rate of change of cost at a production level of x = 3 units is -2/3.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the rate of change of cost at x = 3 units as -2/3, indicating that for every unit increase in production, the cost decreases by 2/3 units.</p>
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<p>We find the rate of change of cost at x = 3 units as -2/3, indicating that for every unit increase in production, the cost decreases by 2/3 units.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y = 6/x.</p>
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<p>Derive the second derivative of the function y = 6/x.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>The first step is to find the first derivative, dy/dx = -6/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-6/x²] Using the power rule, d²y/dx² = 12/x³ Therefore, the second derivative of the function y = 6/x is 12/x³.</p>
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<p>The first step is to find the first derivative, dy/dx = -6/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-6/x²] Using the power rule, d²y/dx² = 12/x³ Therefore, the second derivative of the function y = 6/x is 12/x³.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use a step-by-step process, where we start with the first derivative.</p>
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<p>We use a step-by-step process, where we start with the first derivative.</p>
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<p>Using the power rule, we differentiate -6/x².</p>
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<p>Using the power rule, we differentiate -6/x².</p>
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<p>We then simplify the terms to find the final answer.</p>
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<p>We then simplify the terms to find the final answer.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Prove: d/dx (3/x²) = -6/x³.</p>
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<p>Prove: d/dx (3/x²) = -6/x³.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Let’s start using the power rule: Consider y = 3/x² Rewriting as y = 3·x⁻², To differentiate, we use the power rule: dy/dx = 3·(-2)x⁻³ = -6/x³ Hence proved.</p>
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<p>Let’s start using the power rule: Consider y = 3/x² Rewriting as y = 3·x⁻², To differentiate, we use the power rule: dy/dx = 3·(-2)x⁻³ = -6/x³ Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this step-by-step process, we used the power rule to differentiate the equation.</p>
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<p>In this step-by-step process, we used the power rule to differentiate the equation.</p>
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<p>Then, we replace x⁻² with its derivative to derive the equation.</p>
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<p>Then, we replace x⁻² with its derivative to derive the equation.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<h3>Problem 5</h3>
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<p>Solve: d/dx (6/x²)</p>
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<p>Solve: d/dx (6/x²)</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>To differentiate the function, we rewrite it using exponents: d/dx (6/x²) = d/dx (6·x⁻²) Applying the power rule, = 6·(-2)x⁻³ = -12/x³ Therefore, d/dx (6/x²) = -12/x³.</p>
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<p>To differentiate the function, we rewrite it using exponents: d/dx (6/x²) = d/dx (6·x⁻²) Applying the power rule, = 6·(-2)x⁻³ = -12/x³ Therefore, d/dx (6/x²) = -12/x³.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we differentiate the given function using the power rule.</p>
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<p>In this process, we differentiate the given function using the power rule.</p>
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<p>As a final step, we simplify the equation to obtain the final result.</p>
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<p>As a final step, we simplify the equation to obtain the final result.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h2>FAQs on the Derivative of 6/x</h2>
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<h2>FAQs on the Derivative of 6/x</h2>
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<h3>1.Find the derivative of 6/x.</h3>
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<h3>1.Find the derivative of 6/x.</h3>
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<p>Using the power rule for 6·x⁻¹ gives, d/dx (6/x) = -6/x² (simplified).</p>
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<p>Using the power rule for 6·x⁻¹ gives, d/dx (6/x) = -6/x² (simplified).</p>
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<h3>2.Can we use the derivative of 6/x in real life?</h3>
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<h3>2.Can we use the derivative of 6/x in real life?</h3>
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<p>Yes, we can use the derivative of 6/x in real life in areas such as economics to determine the rate of change of cost with respect to production levels.</p>
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<p>Yes, we can use the derivative of 6/x in real life in areas such as economics to determine the rate of change of cost with respect to production levels.</p>
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<h3>3.Is it possible to take the derivative of 6/x at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of 6/x at the point where x = 0?</h3>
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<p>No, x = 0 is a point where 6/x is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
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<p>No, x = 0 is a point where 6/x is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
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<h3>4.What rule is used to differentiate 6/x²?</h3>
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<h3>4.What rule is used to differentiate 6/x²?</h3>
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<p>We use the power rule to differentiate 6/x² by rewriting it as 6·x⁻², d/dx (6·x⁻²) = -12/x³.</p>
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<p>We use the power rule to differentiate 6/x² by rewriting it as 6·x⁻², d/dx (6·x⁻²) = -12/x³.</p>
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<h3>5.Are the derivatives of 6/x and 6x⁻¹ the same?</h3>
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<h3>5.Are the derivatives of 6/x and 6x⁻¹ the same?</h3>
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<p>Yes, they are the same. The derivative of 6/x is -6/x², which is equivalent to differentiating 6x⁻¹ using the power rule.</p>
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<p>Yes, they are the same. The derivative of 6/x is -6/x², which is equivalent to differentiating 6x⁻¹ using the power rule.</p>
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<h2>Important Glossaries for the Derivative of 6/x</h2>
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<h2>Important Glossaries for the Derivative of 6/x</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x. </li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x. </li>
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</ul><ul><li><strong>Power Rule:</strong>A basic differentiation rule used to find the derivative of a function of the form axⁿ. </li>
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</ul><ul><li><strong>Power Rule:</strong>A basic differentiation rule used to find the derivative of a function of the form axⁿ. </li>
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</ul><ul><li><strong>Undefined Points:</strong>Points where a function does not have a value, often due to division by zero.</li>
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</ul><ul><li><strong>Undefined Points:</strong>Points where a function does not have a value, often due to division by zero.</li>
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</ul><ul><li><strong>Rate of Change:</strong>A measure of how a quantity changes with respect to another variable. </li>
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</ul><ul><li><strong>Rate of Change:</strong>A measure of how a quantity changes with respect to another variable. </li>
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</ul><ul><li><strong>Asymptote:</strong>A line that a curve approaches but never touches or intersects.</li>
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</ul><ul><li><strong>Asymptote:</strong>A line that a curve approaches but never touches or intersects.</li>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<h2>Jaskaran Singh Saluja</h2>
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<h2>Jaskaran Singh Saluja</h2>
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<h3>About the Author</h3>
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<h3>About the Author</h3>
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<p>Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.</p>
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<p>Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.</p>
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<h3>Fun Fact</h3>
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<h3>Fun Fact</h3>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>